9x^2-12x+10=0 eching

$$9 { x }^{ 2 } -12x+10=0$$

x uchun yechish (complex solution)

$$x=\frac{2+\sqrt{6}i}{3}\approx 0,666666667+0,816496581i$$

$$x=\frac{-\sqrt{6}i+2}{3}\approx 0,666666667-0,816496581i$$

Kvadrat tenglamalaridan foydalanish qoidalari

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni kvadrat formulasi bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat formula ikki yechmni taqdim qiladi, biri ± qo'shish bo'lganda, va ikkinchisi ayiruv bo'lganda.

$$9x^{2}-12x+10=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 9 ni a, -12 ni b va 10 ni c bilan almashtiring.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 9\times 10}}{2\times 9}$$

-12 kvadratini chiqarish.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 9\times 10}}{2\times 9}$$

-4 ni 9 marotabaga ko'paytirish.

$$x=\frac{-\left(-12\right)±\sqrt{144-36\times 10}}{2\times 9}$$

-36 ni 10 marotabaga ko'paytirish.

$$x=\frac{-\left(-12\right)±\sqrt{144-360}}{2\times 9}$$

144 ni -360 ga qo'shish.

$$x=\frac{-\left(-12\right)±\sqrt{-216}}{2\times 9}$$

-216 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-12\right)±6\sqrt{6}i}{2\times 9}$$

-12 ning teskarisi 12 ga teng.

$$x=\frac{12±6\sqrt{6}i}{2\times 9}$$

2 ni 9 marotabaga ko'paytirish.

$$x=\frac{12±6\sqrt{6}i}{18}$$

x=\frac{12±6\sqrt{6}i}{18} tenglamasini yeching, bunda ± musbat. 12 ni 6i\sqrt{6} ga qo'shish.

$$x=\frac{12+6\sqrt{6}i}{18}$$

12+6i\sqrt{6} ni 18 ga bo'lish.

$$x=\frac{2+\sqrt{6}i}{3}$$

x=\frac{12±6\sqrt{6}i}{18} tenglamasini yeching, bunda ± manfiy. 12 dan 6i\sqrt{6} ni ayirish.

$$x=\frac{-6\sqrt{6}i+12}{18}$$

12-6i\sqrt{6} ni 18 ga bo'lish.

$$x=\frac{-\sqrt{6}i+2}{3}$$

Tenglama yechildi.

$$x=\frac{2+\sqrt{6}i}{3} x=\frac{-\sqrt{6}i+2}{3}$$

Kvadratni yakunlash bo'yicha qadamlar

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$9x^{2}-12x+10=0$$

Tenglamaning ikkala tarafidan 10 ni ayirish.

$$9x^{2}-12x+10-10=-10$$

O‘zidan 10 ayirilsa 0 qoladi.

$$9x^{2}-12x=-10$$

Ikki tarafini 9 ga bo‘ling.

$$\frac{9x^{2}-12x}{9}=-\frac{10}{9}$$

9 ga bo'lish 9 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{12}{9}\right)x=-\frac{10}{9}$$

\frac{-12}{9} ulushini 3 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}-\frac{4}{3}x=-\frac{10}{9}$$

-\frac{4}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{2}{3} olish uchun. Keyin, -\frac{2}{3} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{4}{3}x+\left(-\frac{2}{3}\right)^{2}=-\frac{10}{9}+\left(-\frac{2}{3}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{2}{3} kvadratini chiqarish.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{-10+4}{9}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{10}{9} ni \frac{4}{9} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=-\frac{2}{3}$$

x^{2}-\frac{4}{3}x+\frac{4}{9} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{2}{3}\right)^{2}=-\frac{2}{3}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{-\frac{2}{3}}$$

Qisqartirish.

$$x-\frac{2}{3}=\frac{\sqrt{6}i}{3} x-\frac{2}{3}=-\frac{\sqrt{6}i}{3}$$

\frac{2}{3} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{2+\sqrt{6}i}{3} x=\frac{-\sqrt{6}i+2}{3}$$